# Evaluate the triple integral $\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$ using spherical coordinates

How to evaluate triple integral $$\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$$ when $$E$$ is bounded by $$x^2+y^2+z^2-x=0$$?

I know that spherical coordinates mean that $$x=r\sin\theta\cos\varphi,\quad y=r\sin\theta\sin\varphi,\quad z=r\cos\theta$$ and this function in spherical coordinates is \begin{align*} &\iiint\limits_E\frac{yzdxdydz}{x^2+y^2+z^2} = \iiint\limits_E\frac{r^2\sin\theta}{r^2\sin^2\theta\cos^2\varphi + r^2\sin^2\theta\sin^2\varphi + r^2\cos^2\theta}drd\theta d\varphi = \\ &\iiint\limits_E\frac{r^2\sin\theta}{r^2(\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta)}drd\theta d\varphi = \iiint\limits_E\frac{\sin\theta}{\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta}drd\theta d\varphi \end{align*} but I don't know how to write $$E$$ as set and convert it to spherical coordinates, and also what happens with this function after conversion. Triple integrals is now topic for me and I have never used spherical coordinates before, so I would be grateful if anyone can help me with this.

• One thing to note: from the identities you were given for $x,y,z$, you can prove that $$x^2 + y^2 + z^2 = r^2$$ This will make your integration significantly easier. Not really sure how to help you on setting the bounds of the integral, though. Sometimes it's best to just graph it and see if you can reason it geometrically, but maybe someone will offer a more proper answer towards that end. Commented Nov 26, 2020 at 8:36
• I would use cylindrical polar coordinates, symmetric about the $x$ axis, so $y=r\cos\theta, z=r\sin\theta$ Commented Nov 26, 2020 at 9:07
• The command of Mathematica Integrate[ y*z/(x^2 + y^2 + z^2), {x, y, z} \[Element] ImplicitRegion[x^2 + y^2 + z^2 - x <= 0, {x, y, z}]] produces $0$. Commented Nov 26, 2020 at 10:10

The set $$E$$ is a ball of radius $${1\over2}$$, centered at $$\left({1\over2},0,0\right)$$. The integrand $$f(x,y,z):={yz\over x^2+y^2+z^2}$$ satisfies $$f(x,-y,z)\equiv-f(x,y,z)$$. This means that $$f$$ is odd with respect to the symmetry plane $$y=0$$ of $$E$$. It is then obvious that the requested integral has value $$0$$.

Using the spherical coordinates (with the convention you used) we have, $$yz= r^2\sin\theta\cos\theta\sin\varphi,$$ $$x^2+y^2+z^2 = r^2$$ $$\,dx\,dy\,dz = r^2\sin\theta\,dr\,d\varphi\,d\theta.$$ Also, the region $$E$$ is bounded by $$x^2+y^2+z^2-x = 0$$, which is a sphere centered at $$(\tfrac{1}{2},0,0)$$ with radius $$\tfrac{1}{2}$$.

The inside of this sphere is given by $$x^2+y^2+z^2-x \le 0$$, i.e. $$r^2-r\sin\theta\cos\varphi \le 0$$. Since $$r \ge 0$$, this simplifies to $$0 \le r \le \sin\theta \cos\varphi$$ or $$r = 0$$ (which is just one point).

In order for $$0 \le r \le \sin\theta\cos\varphi$$ to be a nontrivial range, we need $$\sin\theta\cos\varphi \ge 0$$. Since $$\sin\theta \ge 0$$ for all $$\theta \in [0,\pi]$$, we only need to restrict the bounds of $$\varphi$$ to be over the range where $$\cos\varphi \ge 0$$. We can do this either by $$\varphi \in [0,\tfrac{\pi}{2}] \cup [\tfrac{3\pi}{2},2\pi]$$ or $$\varphi \in [-\tfrac{\pi}{2},\tfrac{\pi}{2}]$$. The second is simpler, so we will use that instead.

So the integral becomes $$\iiint\limits_{E}\dfrac{yz \,dx\,dy\,dz}{x^2+y^2+z^2} = \int_{-\pi/2}^{\pi/2}\int_{0}^{\pi}\int_{0}^{\sin\theta\cos\varphi}r^2\sin^2\theta\cos\theta\sin\varphi\,dr\,d\theta\,d\varphi.$$

• ^Thanks for catching that. Commented Nov 26, 2020 at 21:36
• You are welcome! Commented Nov 26, 2020 at 21:37

$$x^2+y^2+z^2-x=0 \implies (x-\frac{1}{2})^2 + y^2 + z^2 = (\frac{1}{2})^2$$

So it is a sphere with radius $$\frac{1}{2}$$ centered at $$(\frac{1}{2},0,0)$$.

In spherical coordinates, $$x = \rho \cos \theta \sin \phi, \, y = \rho \sin \theta \sin \phi, z = \rho \cos \phi \,$$ where $$\theta$$ is the azimuthal angle and $$\phi$$ is the polar angle (just opposite of your convention).

$$x^2 + y^2 + z^2 = \rho^2$$ so substituting in the equation of our sphere,

$$\rho^2 = \rho \cos \theta \sin \phi \implies \rho = \cos \theta \sin \phi$$.

Now the part that you have to be careful about is the bounds of azimuthal angle. Please consider a circle in XY plane with center at $$(\frac{1}{2},0)$$ and radius of $$\frac{1}{2}$$ and equation of $$r = \cos \theta$$. $$\theta$$ varies betweeen -$$\frac{\pi}{2}$$ and $$\frac{\pi}{2}$$ (range of $$\pi$$ instead of $$2 \pi$$ for a sphere centered at the origin). It is same for this sphere.

So the integral becomes

$$I = \displaystyle \int_{-\pi/2}^{\pi/2} \int_{0}^{\pi} \int_{0}^{\cos\theta \sin\phi} \rho^2 \sin \theta \sin^2 \phi \cos \phi \, d\rho \, d\phi \, d\theta$$

And based on the given function, its integral above $$XY$$ plane and below will cancel each other.

If I was integrating over the part of the sphere only in the first octant, the integral will be

$$I = \displaystyle \int_{0}^{\pi/2} \int_{0}^{\pi/2} \int_{0}^{\cos\theta \sin\phi} \rho^2 \sin \theta \sin^2 \phi \cos \phi \, d\rho \, d\phi \, d\theta = \frac{1}{72}$$